Answer :
Answer: The verification is done below.
Step-by-step explanation: We are given to show that the function [tex]f(x)=4e^{3x}-3e^{-x}[/tex] is a solution to the following differential equation :
[tex]y^{\prime\prime}-y^{\prime}-3y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
If y = f(x), then we see that
[tex]y^\prime=\dfrac{d}{dx}f(x)=\dfrac{d}{dx}(4e^{3x}-3e^{-x})=12e^{3x}+3e^{-x},\\\\\\y^{\prime\prime}=\dfrac{d}{dx}(12e^{3x}+3e^{-x})=36e^{3x}-3e^{-x}.[/tex]
Therefore, we get
[tex]L.H.S.\\\\\\=y^{\prime\prime}-2y^{\prime}-3y\\\\\\=(36e^{3x}-3e^{-x})-2(12e^{3x}+3e^{-x})-3(4e^{3x}-3e^{-x})\\\\\\=36e^{3x}-3e^{-x}-24e^{3x}+6e^{-x}-12e^{3x}+9e^{-x}\\\\=0\\\\=R.H.S.[/tex]
Thus, the function [tex]f(x)=4e^{3x}-3e^{-x}[/tex] is a solution to the given differential equation.
Hence showed.